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The eigen vector $(s)$ of the matrix $$\begin{bmatrix} 0 &0 &\alpha\\ 0 &0 &0\\ 0 &0 &0 \end{bmatrix},\alpha \neq 0$$ is (are)

1. $(0,0,\alpha)$
2. $(\alpha,0,0)$
3. $(0,0,1)$
4. $(0,\alpha,0)$

I believe the answer should be only B. It is because if we look at the matrix it shows the transformed 3D space into a line where the z^ vector is transformed into a line replicating x axis. In that scenario the only vector that remains unchanged is (alpha, 0, 0) i.e option B.
could anyone give the complete procedure ?
Answer is b only why obtion d is consider here ?
Why d is included in answer?

Here we got eign values as 0,0,0.

Now to get eign vector take  λ=0 in

( A-λI) X =0 or AX=0 -----(i)

here X is eign vector in form of column matrix [X1,X2,X3].

If we multiply as per (i) we get equation as

X1*0+X2*0+ aX3=0

Now for LHS to be 0 , X3 must be 0 . Here X1 and X2 can be assumed any value . As per the given option a and c are wrong.hence b and d satisfy the case
what about (alpha,alpha,0) vectors? will they be eigen vectors too.I think they are.
Bind blowing explanation satyajeet sir
Yes they will be eigen [email protected] gow

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Since, the given matrix is an upper triangular one, all eigenvalues are 0. And hence $A-\lambda I=A$

$\begin{bmatrix} 0 &0 &\alpha \\ 0&0 &0 \\ 0& 0& 0 \end{bmatrix}\begin{bmatrix} x_1\\ x_2\\ x_3 \end{bmatrix}=\begin{bmatrix} 0\\ 0\\ 0 \end{bmatrix}$

what $x_1,x_2,x_3$ are suitable?

which means $x_1$times column1+$x_2$times column2+$x_3$ times column3=Zero Vector

Look carefully,if you take any linear combination of columns 1 and 2 and don't take column 3(means $x_3=0$), the result will be a zero vector.

So, if we have our X=$\begin{bmatrix} \beta\\ \gamma\\ 0 \end{bmatrix}$ where $\beta, \gamma$ are any scalar quantity,

but $x_3$ must be necessarily zero to get zero vector.

Hence, only option (B) and (D) satisfy.

The second part of the answer is difficult for me to grasp.
Which one?
From the statement "look carefully"
ohh that one. See from this example.

Suppose you have a system like

$\begin{bmatrix} 1 &3 &6 \\ 7 & 5& 9\\ 3 &8 & 2 \end{bmatrix}\begin{bmatrix} x_1\\ x_2\\ x_3 \end{bmatrix}$

this is equivalent to

$x_1\begin{bmatrix} 1\\ 7\\ 3 \end{bmatrix}+x_2\begin{bmatrix} 3\\ 5\\ 8 \end{bmatrix}+x_3\begin{bmatrix} 6\\ 9\\ 2 \end{bmatrix}$

In the question given, you have to produce a zero vector as result. What suitable combinations of $x_1,x_2,x_3$ can you take?
Good Explanation Sir ;-)
@Ayush sir, it will be 0,0,0 right?
Given matrix is an upper triangular matrix. Therefore diagonal elements of A are eigen values of A, i.e. λ=0,0,0.

Consider the characteristic equation $(A-λI)x=0$

$\Rightarrow$ $\begin{bmatrix} 0-\lambda & 0 & \alpha \\ 0 & 0-\lambda & 0\\ 0 & 0 & 0-\lambda \end{bmatrix}$ $\begin{bmatrix}x_1\\ x_2\\ x_3 \end{bmatrix}$ = $\begin{bmatrix} 0\\ 0\\ 0 \end{bmatrix}$

To find the eigen vector of A corresponding to $\lambda$
Put the eigen value of A in the characteristic equation

i.e. $\begin{bmatrix} 0 & 0 & \alpha \\ 0 & 0 & 0\\ 0 & 0 & 0 \end{bmatrix}$ $\begin{bmatrix}x_1\\ x_2\\ x_3 \end{bmatrix}$ = $\begin{bmatrix} 0\\ 0\\ 0 \end{bmatrix}$

Now we have

$\alpha$$x_3=0 \Rightarrow$$x_3$ = 0

Therefore, any non zero vector with z-component as zero is an eigen vector.
Hence in the given options $B$ and $D$ are correct.
by

### 1 comment

Thank you for explaining :)
Answer: $B, D$

Eigen values are: $0,0,0$

The eigen vector should satisfy the equation: $\alpha$z $=$ $0$

yes, I also think ans should be option B only.
how b and d?
here by normal approach we get in eigen vector x3=0

So, answer will be b) and d)
^ Will $-\lambda^3 = 0$ not be the characterastic equations for this matrix?
So how third value came $\alpha$?
yes (B) and (D) should be correct option!

here we have the Basic variable as "z" and free variable as "x" and "y".

So, keeping x = t and y = s, and z = $\alpha$

Then vector corresponding to x is $[1, 0, 0]^T$

And vector corresponding to y is $[0, 1, 0]^T$

Then why it is not $[1, 0, 0]^T$ And $[0, 1, 0]^T$ instead of $[\alpha, 0 , 0]^T$ And $[0, \alpha, 0]^T$?

$\alpha$ or 1 doesnot matter

atleast u have to give answer, from options given
But it should be $[1,0,0]^T$ And $[0,1,0]^T$ even no option matches. ryt??
if B & D are true then A should also be true please check once?
Ok

Guys this is an easy question, so let us not make it complex.

As most of the people have already explained that $|A-\lambda I|*eigenvector =\begin{bmatrix} 0 &0 &\alpha \\ 0 & 0 &0 \\ 0& 0 & 0 \end{bmatrix}$$\begin{bmatrix} x_{1}\\ x_{2}\\ x_{3} \end{bmatrix}$

Now see that we are getting $0*x_{1} + 0*x_{2} + \alpha *x_{3}=0$ as the only useful equation.

Now put whatever the value of $x_{1}$ ,$x_{2}$ and $x_{3}$, we will get our eigenvectors, EXCEPT the cases where $\alpha$ has to be zero because in question it clearly mentioned that $\alpha \neq 0$. This condition will only fail when $x_{3}\neq 0.$ And therefore now you can match the options very easily.

by

clearly matrix is upper triangular matrix and for upper triangular ,lower triangular matrix ,diagonal matrix and scaler matrix ,its diagonal element is eigen values.so here eigen value=0,0,0.

now we have to find eigen vector of this matrix (A-λ)X=0  where λ=0 so we get A matrix  [ 0 0 α] has rank=1  means we get two free variable (n-r)

x1= β x2=γ x3=0 then eigen vector  (x1,x2,x3)=(β,γ,0)= β(1,0,0)+γ(0,1,0) so  B and  D  are  right answer .

https://youtu.be/VqP2tREMvt0

Check @32:57

After solving the characteristic equation $\det(A – \lambda I_n)=0$ we get $\lambda=0,0,0$. Using this $\lambda=0$ in $A-\lambda I_n$ we have:

Here $C_1$ and $C_2$ are any real. We may even choose them as $C_1=C_2=\alpha$ to get options $B$ and $D$.

by